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3 years ago

For each equation, determine if x=2 is a solution.

Mathematics

1 answer:

den301095 [7]3 years ago

4 0

Answer:

-2(2-4)=4

-3.8x= -7.6

4(2-1)-3(2-2)

Step-by-step explanation:

Negative two times two equals negative four,negative two times negative four equals eight,negative four plus eight equals four.

Two times negative one point six, two times negative three six, negative six plus negative one point six equals negative seven point six.

Four times two and negative one equals four, negative three times two and negative two equals zero, four plus zero equals four.

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4 years ago

tims math homework is on five consecutive pages in the math textbook, and the sum of those page numbers is 630. what is the page

Brrunno [24]

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Step-by-step explanation:

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3 years ago

PLEASE HELP ME GUYS OR I WONT PASS this calculus!!!!

KonstantinChe [14]

Answer:

b. $\displaystyle \frac{1}{2}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Functions
Function Notation
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle H(x) = \sqrt[3]{F(x)}$

Step 2: Differentiate

Rewrite function [Exponential Rule - Root Rewrite]: $\displaystyle H(x) = [F(x)]^\bigg{\frac{1}{3}}$
Chain Rule: $\displaystyle H'(x) = \frac{d}{dx} \bigg[ [F(x)]^\bigg{\frac{1}{3}} \bigg] \cdot \frac{d}{dx}[F(x)]$
Basic Power Rule: $\displaystyle H'(x) = \frac{1}{3}[F(x)]^\bigg{\frac{1}{3} - 1} \cdot F'(x)$
Simplify: $\displaystyle H'(x) = \frac{F'(x)}{3}[F(x)]^\bigg{\frac{-2}{3}}$
Rewrite [Exponential Rule - Rewrite]: $\displaystyle H'(x) = \frac{F'(x)}{3[F(x)]^\bigg{\frac{2}{3}}}$

Step 3: Evaluate

Substitute in x [Derivative]: $\displaystyle H'(5) = \frac{F'(5)}{3[F(5)]^\bigg{\frac{2}{3}}}$
Substitute in function values: $\displaystyle H'(5) = \frac{6}{3(8)^\bigg{\frac{2}{3}}}$
Exponents: $\displaystyle H'(5) = \frac{6}{3(4)}$
Multiply: $\displaystyle H'(5) = \frac{6}{12}$
Simplify: $\displaystyle H'(5) = \frac{1}{2}$